Let N be the number of ordered pairs $(a,b)$ of natural numbers such that $a <b$ and the H.M of $a$ and $b$ is $2010$. Then last digit of N is?

Solution 1:

My hint in comments:

Let $c=a−1005$ so $b=\frac{1005^2}c+1005$

Your response:

I think $N$ is $13$. $c$ is a factor of $1005^2$ and less than $1005$. So there will be $13$ such choices for $c$

Indeed $13$ is correct.

$1005=3\times5\times 67$ so $1005^2$ has $3^{2+1}=27$ factors of which $13$ are less than $1005$, $1$ is $1005$, and $13$ are greater than $1005$.

The $13$ possibilities are in fact

    c   a       b   
    1 1006 1011030
    3 1008  337680
    5 1010  203010
    9 1014  113230
   15 1020   68340
   25 1030   41406
   45 1050   23450
   67 1072   16080
   75 1080   14472
  201 1206    6030
  225 1230    5494
  335 1340    4020
  603 1608    2680